Another kinematics problem

A rifle is aimed horizontally at the center of a large target 200 m away. The initial speed of the bullet is 500 m/s. Where does the bullet strike the target? To hit the center of the target, the barrel must be at an angle above the line of sight. Find the angle of elecation of the barrel.

So far I know that

in the x-direction:

V0 = 500 m/s
x= 200 m

in the y-direction:
V0= 0 m/2
a = 9.81 m/s^2

and I know that they have time in common. But I don't really know where to go from here. Do we even have enough information to solve the problem?

You have 2 exercises here:
1) The rifle is aimed horizontally
You have listed the correct quantities you must use here.
2) What angle must you aim the rifle along in order to hit the center of the target?

1):
Let the origin be the muzzle of the rifle (i.e, x=0,y=0)
(This means, in particular, that the ground level has some unspecified, negative value)
a)What time does it take for the bullet to travel 200 meters horizontally?
b) After that time, how far down from y=0 has the bullet fallen?

2)
The speed of the bullet is [tex]V_{0}=500[/tex]
With a non-zero angle to the vertical, we have for the initial horizontal velocity:
[tex]V_{0,x}=500\cos\theta[/tex]
The initial vertical velocity satisfies:
[tex]V_{0,y}=500\sin\theta[/tex]

(Note that in the particular case [tex]\theta=0[/tex] we retrieve the relations for a horizontally aimed rifle)

a) What is now the time the bullet takes to traverse 200 meters in the horizontal direction?
b) Solve now for the final vertical position as a function of the initial angle.
You want to find the angle such that [tex]y_{final}=0[/tex]